Question

Sum of the series $\sum _{r=0}^n(-1{)}^r{}^n{C}_r\cdot [i^{5r}+i^{6r}+i^{7r}+i^{8r}],$ is

• A. $2^n$
• B. $2^{\frac{n}{2}+1}$
• C. $n^n+2^{\frac{n}{2}+1}$
• D. $2^n+2^{\frac{n}{2}+1}\cdot \cos \frac{n\pi }{4}$

Answer 1

The correct option is D $2^n+2^{\frac{n}{2}+1}\cdot \cos \frac{n\pi }{4}$

We know that,
$\sum _{r=0}^n(-1{)}^r{}^n{C}_r{x}^r=(1-x{)}^n$

$\therefore \sum _{r=0}^n(-1{)}^r{}^n{C}_r{i}^{5r}=(1-i^5{)}^n=(1-i{)}^n$

Similarly,
$\sum _{r=0}^n(-1{)}^r{}^n{C}_r{i}^{6r}=(1-i^6{)}^n=(1-(-1){)}^n$

$\sum _{r=0}^n(-1{)}^r{}^n{C}_r{i}^{7r}=(1-i^7{)}^n=(1-(-i){)}^n$

$\sum _{r=0}^n(-1{)}^r{}^n{C}_r{i}^{8r}=(1-i^8{)}^n=(1-1{)}^n$

Hence,
$\sum _{r=0}^n(-1{)}^r{}^n{C}_r\cdot [i^{5r}+i^{6r}+i^{7r}+i^{8r}]=(1-i{)}^n+(1-(-1){)}^n+(1-(-i){)}^n+(1-1{)}^n$
$=(1-i{)}^n+2^n+(1+i{)}^n+0=\left(\sqrt{2}{)}^n\right(e^{-i\pi /4}{)}^n+2^n+\left(\sqrt{2}{)}^n\right(e^{i\pi /4}{)}^n=\left(\sqrt{2}{)}^n\right(e^{in\pi /4}+e^{-in\pi /4})+2^n=(\sqrt{2}{)}^n\left(2\cos \frac{n\pi }{4}\right)+2^n=2^n+2^{\frac{n}{2}+1}\cdot \cos \frac{n\pi }{4}$

$\text{Alternate Solution}$
$\sum _{r=0}^n(-1{)}^r{}^n{C}_r\cdot [i^{5r}+i^{6r}+i^{7r}+i^{8r}]$
Let $n=1$
$=\sum _{r=0}^1(-1{)}^r{}^1{C}_r\cdot [i^{5r}+i^{6r}+i^{7r}+i^{8r}]$
$=(-1{)}^0{}^1{C}_0\cdot [i^0+i^0+i^0+i^0]+(-1{)}^1{}^1{C}_1\cdot [i^5+i^6+i^7+i^8]$
$=4+(-1)[i-1-i+1]$
$=4+0=4$
option $2^n+2^{\frac{n}{2}+1}\cdot \cos \frac{n\pi }{4}$ gives $4$ at $n=1$